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I have to find for which primes $p$, the equation $x^2+y^2=3z^2$ has a rational point in $\mathbb{Q}_p$.

According to my notes:

Obviously, $\forall p \in \mathbb{P}, p \nmid 2 \cdot 3$, there is a rational solution in $\mathbb{Q}_p$.

But,why is it obvious that the equation has a rational solution in $\mathbb{Q}_p,\forall p \in \mathbb{P}, p \nmid 2 \cdot 3 $ ?

Then, it is shown that the equation has no non-trivial solutions in $\mathbb{Q}_3$.

Now, I have to show that the equation has no solution in $\mathbb{Q}_2$.

But, why doesn't it have a solution in $\mathbb{Q}_2$?

We have $x^2+y^2-z^2 \equiv 0 \pmod 2$. Isn't $(x,y,z)=(1,1,0)$ a solution?

Or am I wrong?

Bart Michels
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evinda
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    With $(x,y,z)=(1,1,0)$ we get $x^2+y^2=2$ and $3z^2=0$. $\Bbb{Q}_2$ has characteristic zero, so $2\neq0$ in there. – Jyrki Lahtonen Nov 03 '14 at 19:43
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    Yeah, you seem confused about the nature of $\mathbb Q_p$. $1+1\neq 0$ in $\mathbb Q_2$. On the other hand, $1+1=0$ in $\mathbb F_2\cong \mathbb Z/2\mathbb Z$, which is what you've shown. – Thomas Andrews Nov 03 '14 at 19:45
  • So, when an equation is not equivalent to $0$, modulo $p$, we know that the equation has no solution in $\mathbb{Q}$ and if it is equivalent to $0$, modulo $5$, we have to check if the equality holds in $\mathbb{Q}$ ? Or have I understood it wrong? – evinda Nov 03 '14 at 19:48
  • what does it mean for an equation to be "equivalent to zero?" – Thomas Andrews Nov 03 '14 at 19:51
  • If it is equivalent to $0 \pmod p$, it is a multiple of $p$, right? – evinda Nov 03 '14 at 19:52
  • I have no idea what you are trying to say, is what I mean. That was not a Socratic question. Why are we suddenly talking about $5$? – Thomas Andrews Nov 03 '14 at 19:55
  • Give me an example of an "equation equivalent to zero modulo $p$." – Thomas Andrews Nov 03 '14 at 19:56
  • A mistranslation. The correct term is congruent to. As in a number can be congruent to another (modulo something). An equation can be equivalent to another equation, but not to a number. – Jyrki Lahtonen Nov 03 '14 at 19:56
  • @JyrkiLahtonen That doesn't clear anything up. Example. – Thomas Andrews Nov 03 '14 at 19:58
  • @Thomas: Evinda's comment a few lines up. "If it is equivalent to $0\pmod p$, it is a multiple of $p$, right?" It may IMHO be a mistranslation of "If it is congruent to $0\pmod p$, it is a multiple of $p$, right?" – Jyrki Lahtonen Nov 03 '14 at 20:03
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    Evinda, the phrase "rational solution in $\Bbb{Q}_p$" is also misleading. Not all elements of $\Bbb{Q}_p$ are rational. $\Bbb{Q}_p$ is uncountable! – Jyrki Lahtonen Nov 03 '14 at 20:14
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    If what is congruent to zero? @JyrkiLahtonen The problem is the vagueness. Is it the equation - that is, it is one where all the terms of the equation are divisible by $p$ - or is he saying there is some solution to the equation modulo $p$? I really have no idea. – Thomas Andrews Nov 03 '14 at 20:34

1 Answers1

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In $\Bbb{Q}_2$ you often work modulo $8$, and that is the first thing to try here, too. Assume that a non-trivial solution would exist. A triple $(x,y,z)$ is a solution if and only if $(2x,2y,2z)$ is, so by iterating this enough many times we can assume that all of $(x,y,z)$ are $2$-adic integers, and at least one of them is a $2$-adic unit.

After that we can reduce modulo eight, replace the equation with a congruence modulo $8$, and assume that $x,y,z\in\{0,1,2,3,4,5,6,7\}$ with at least one of them odd. But for all odd integers $n$ we have $n^2\equiv1\pmod 8$. OTOH even integers have square $\equiv0$ or $\equiv 4\pmod 8$. So $x^2, y^2, z^2$ all come from the set $\{0,1,4\}\pmod 8$, and at least one of them is $=1$. It is straightforward to verify that no combo works modulo $8$.

When $p>3$ we can arrange to use Hensel's lemma. First I set $z=1$ and prove that the equation $$x^2+y^2=3$$ has a solution $(x,y)\in\Bbb{F}_p^2$. This a standard counting argument. There are $(p+1)/2$ squares in $\Bbb{F}_p$. Therefore the sets $$ A=\{x^2\mid x\in\Bbb{F}_p\}\qquad\text{and}\qquad B=\{3-y^2\mid y\in\Bbb{F}_p\} $$ both have $(p+1)/2$ elements. Because there are only $p$ elements in $\Bbb{F}_p$ the intersection $A\cap B$ is non-empty. If $u\in A\cap B$, then $x^2=u=3-y^2$ for some choices $x,y\in\Bbb{F}_p$. But for those $x,y$ we have $x^2+y^2=3$ as claimed.

Clearly either $x$ or $y$ is non-zero (here we need $p\neq3$). Their roles are interchangeable, so w.l.o.g. we can assume that $x\neq0$. Let us now view these $x,y$ as an integer. Consider the polynomial $$ F(T)=T^2-(3-y^2)\in\Bbb{Z}[T]. $$ We have $F(x)\equiv 0\pmod p$. As $F'(T)=2T$ and $x\neq 0_{\Bbb{F}_p}$ we have $F'(x)\not\equiv0\pmod p$ (here we need $p\neq2$). Hensel's Lemma thus promises us a $p$-adic integer $\alpha$ such that $F(\alpha)=0$.

This means that $(x=\alpha,y,z=1)$ is a solution of $x^2+y^2=3z^2$.

Jyrki Lahtonen
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  • Jyrki Lahtonen Having shown that $x^2+y^2-3z^2 \not\equiv 0 \pmod 8$, does this also mean that the equation has no solution in $\mathbb{Q}_2$ ? – evinda Nov 03 '14 at 20:21
  • Yes. The scaling means that if there is a solution in $\Bbb{Q}_2$ then there would be one with $x,y,z\in\Bbb{Z}_2$. Therefore it would also be a solution in $\Bbb{Z}_2/8\Bbb{Z}_2\cong \Bbb{Z}/8\Bbb{Z}.$ – Jyrki Lahtonen Nov 03 '14 at 20:23
  • If there wouldn't be any solution in $\mathbb{Q}_2$, we would conclude that there is also no solution in $\mathbb{Q}_8$, right? $$$$ But we have shown that there is no solution in $\mathbb{Q}_8$. Do we conclude from that, that there is no solution in $\mathbb{Q}_2$ ? – evinda Nov 03 '14 at 20:26
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    There is no $\Bbb{Q}_8$? What do you mean? – Jyrki Lahtonen Nov 03 '14 at 20:32
  • We have shown that there is no solution $\pmod 8$. Does this also mean that there is also no solution $\pmod 2$ ? – evinda Nov 03 '14 at 20:34
  • No. There are solutions modulo $2$. I try to explain a bit more. Wait a minute. – Jyrki Lahtonen Nov 03 '14 at 20:35
  • But how have we shown this? :/ – evinda Nov 03 '14 at 20:35
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    You have to be specific, @evinda. "How have we shown this?" What is "this?" If an equation has a solution in the $2$-adic integers ($\mathbb Z_2$, different from $\mathbb Z/2\mathbb Z$,) then it has a solution modulo $2^n$ for any $n>0$. Your usage of the term $\mathbb Q_8$ indicates you really don't understand what the $2$-adic numbers are. – Thomas Andrews Nov 03 '14 at 20:38
  • We have the field of $p$-adic numbers $\Bbb{Q}_p$. It contains as a subfield the ordinary rationals $\Bbb{Q}$. It also has an interesting infinite subring $\Bbb{Z}_p$ of $p$-adic integers. This subring contains a copy of the usual integers $\Bbb{Z}$ as an even smaller (but still infinite) ring. There is a theorem telling us that the (finite) quotient ring $\Bbb{Z}_p/p^k\Bbb{Z}_p$ is isomorphic to the familiar residue class ring of integers $\Bbb{Z}/p^k\Bbb{Z}$. – Jyrki Lahtonen Nov 03 '14 at 20:40
  • But I agree with Thomas. The source of your misunderstandings lies deeper. I think you should go talk with your teacher for a few hours. It doesn't look like you are at all familiar with the interplay of these rings. Your preceding questions on $p$-adics show similar symptoms. I'm afraid it doesn't look good. – Jyrki Lahtonen Nov 03 '14 at 20:43
  • I haven't understood it.. Having found that there is no solution $\pmod 8$, how do we conclude that there is also no solution $\pmod 2$?????? – evinda Nov 03 '14 at 20:48
  • (cont'd) With $p=2$, $k=3$ this means that there is a surjective homomorphism $f$ from the infinite ring $\Bbb{Z}_2$ to the finite ring $\Bbb{Z}/8\Bbb{Z}$. If $(x,y,z)$ were a solution in $\Bbb{Z}_2$, then $(f(x),f(y),f(z))$ would be a solution in $\Bbb{Z}/8\Bbb{Z}$, but we showed that no such solution exists. Sorry about reposting this comment. I fixed a couple of language mistakes, but the 5-minute edit window had expired. – Jyrki Lahtonen Nov 03 '14 at 20:49
  • There ARE solutions modulo two, you presented one to start with. But a solution modulo $2$ is not a $2$-adic solution. – Jyrki Lahtonen Nov 03 '14 at 20:51
  • This idea is similar to proving the non-existence of a solution over integers by showing that no solution w.r.t. some modulus exists. It depends heavily on that isomorphism of quotient rings. – Jyrki Lahtonen Nov 03 '14 at 21:18
  • @JyrkiLahtonen Aren't there $\frac{p-1}{2}$ quadratic residues modulo $p$ ? Or am I wrong? – evinda Nov 06 '14 at 19:26
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    Include zero! ${}$ Can be use here unless $x_0=y_0$. Helps counting but necessitates that extra bit of care. – Jyrki Lahtonen Nov 06 '14 at 21:09
  • @JyrkiLahtonen Does this mean that, no matter at which equatio we look, there is a solution in $\mathbb{Q}_p$, for $p \neq 2$ and $p \nmid $ the coefficients of the equation? – evinda Dec 07 '14 at 14:53
  • @JyrkiLahtonen Also, with $\mathbb{F}_p$ do you mean $\mathbb{Q}_p$ ? – evinda Dec 07 '14 at 14:55
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    Yes. A quadratic homogeneous equation of the form $ax^2+by^2=cz^2$ has non-trivial solutions in $\Bbb{Q}_p$ whenever $p$ is an odd prime not dividing any of $a,b,c$. No. I mean $\Bbb{F}_p=\Bbb{Z}/p\Bbb{Z}$ = the field of $p$ elements. Hensel's lemma is about "lifting" a solution in $\Bbb{F}_p$ to a solution in $\Bbb{Z}_p\subset \Bbb{Q}_p$. – Jyrki Lahtonen Dec 07 '14 at 16:34
  • @JyrkiLahtonen I understand... If we have the equation $3x^2+5y^2-7=0$, do we consider the sets $A={ 3x^2| x \in \mathbb{Z}_p}$ and $B={7-5y^2| y \in \mathbb{Z}_p }$ ? – evinda Dec 08 '14 at 18:56
  • No. $\Bbb{Z}_p$ is an uncountable set, so no counting argument works there. Consider the finite sets $A={3x^2\mid x\in \Bbb{F}_p}$ and $B={7-5y^2\mid y\in\Bbb{F}_p}$. Both are subsets of $(p+1)/2$ numbers of the set $\Bbb{F}_p={\overline{0},\overline{1},\ldots,\overline{p-1}}$ of $p$ elements. Have you by now absorbed that in the context of $p$-adics, $\Bbb{Z}_p$ and $\Bbb{Z}/p\Bbb{Z}=\Bbb{F}_p$ mean different objcets? – Jyrki Lahtonen Dec 09 '14 at 16:31